Erdős Problem #186
Resolved in some way other than a straight proof or disproof — usual for open-ended questions, or for multi-part problems whose parts went different ways.
What the community database records, not our verdict. erdosproblems.com no longer shows statuses (why).
A finite set \(A\) of natural numbers is *non-averaging* if no \(a \in A\) is the arithmetic mean of at least two other elements of \(A\): there is no \(S \subseteq A \setminus \{a\}\) with \(|S| \geq 2\) and \(\sum_{x \in S} x = |S| \cdot a\). This agrees with [PhZa24] ("a nonempty subset of \(A\) not containing \(a\)"), since a single element other than \(a\) never has mean \(a\).
\(F(N)\) is the maximal size of a non-averaging subset of \(\{1, \dots, N\}\).
Attempts
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